Chapter II: Continuum Theory of Dislocations
2.3 Straight Dislocations
Antiplane strain
We now consider an isotropic solid in a state of antiplane strain. In this case, two components of the displacement field are identically zero, with the nonzero component depending only on the position components along which the displace- ments are zero. For instance, for a state of antiplane strain along the x3 axis, the displacement field is of the form
u1 ≡0, u2 ≡0, u3 =u3(x1, x2). (2.21) Consequently, all strain components are zero except for ε13 and ε23 which in this case are given respectively by
ε13= 1
2u3,1 and ε23= 1
2u3,2. (2.22)
Accordingly, the nonzero components of stress are
σ13 = 2µε13=µu3,1 and σ23= 2µε23 =µu3,2. (2.23) Hence, assuming no body forces in the solid, equilibrium equation (2.5) simplifies to
∇2u3 = 0. (2.24)
Thus, the antiplane strain problem reduces to solving Laplace’s equation for u3 together with the relevant boundary conditions.
Figure 2.4: Screw dislocation along the positive x3 axis in a cartesian coordinate system.
antiplane strain as discussed in Section 2.2. Furthermore, u3 is only a function of the angle θ as shown in Figure 2.4 and increases uniformly from zero to b as θ increases from0 to 2π. The simplest form of u3 satisfying the above requirement is [18–20, 26, 35, 42, 44]
u3 = b
2πθ . (2.25)
In cartesian coordinates, this becomes u3 = b
2π tan−1 x2
x1
. (2.26)
One can easily verify that∇2u3 = 0 as required by Equation (2.24). The nonzero components of the strain tensor follow from (2.22) as
ε13 =− b 4π
x2
x21+x22 , (2.27a)
ε23 = b 4π
x1
x21+x22 . (2.27b)
Using Equation (2.23), we obtain the corresponding nonzero stresses as σ13 =−µb
2π x2
x21+x22 , (2.28a)
σ23 = µb 2π
x1
x21+x22 . (2.28b)
The above derivation can also be done in polar coordinates, in which case Equations (2.26), (2.27), and (2.28) take the much simpler form
uz = bθ
2π, (2.29a)
εθz =εzθ = b
4πr, (2.29b)
σθz =σzθ = µb
2πr, (2.29c)
with all other components of displacement, stress, and strain being zero.
• Strain energy
We first notice that as r → 0, the strain εθz and stress σθz diverge to infinity.
Since solids cannot withstand infinite stresses, a hole of radius r0 must be inserted inside the infinite cylinder of Figure 2.4, the result of which is shown in Figure 2.5. However, since real crystals are not hollow, as we approach the center of the dislocation, linear elasticity theory ceases to be valid and a non-linear, atomistic model must be used instead [18, 26, 42]. This region around the dislocation where linear elasticity ceases to be valid is called thecore of the dislocation. It is common to assume a value of r0 on the order of the Burgers vector magnitudeb [18, 26, 44]. Therefore, in computing the strain energy of the screw dislocation, we will focus on the region outside this core and consider the hollow cylinder shown in Figure 2.5.
The strain energy per unit length inside the infinite hollow cylinder can be easily computed using polar coordinates. Using Equations (2.13), (2.29b), and (2.29c), it is found to be
W L =
ˆ R r0
ˆ 2π 0
1
2(σθzεθz+σzθεzθ)rdθdr
= ˆ R
r0
σθzεθz2πrdr
= µb2 4π ln
R r0
.
(2.30)
As excepted, W/Ldiverges as r0 →0. However, it also diverges as R→ ∞. As a result, one cannot associate with the dislocation a definite characteristic energy.
In fact, the energy depends on the size of the crystal. For a single dislocation,R is typically taken to be the distance from the dislocation to the nearest free surface.
For a crsytal containing many dislocations of both signs, R is taken to be half the average distance between the dislocations [26, 42].
Figure 2.5: Screw dislocation along the positivex3 axis in a hollow cylinder used to compute the strain energy of the dislocation.
Edge dislocations
We recall that in an edge dislocation, the Burgers vector is perpendicular to the dislocation line. In deriving the corresponding elastic field, we’ll make use of Figure 2.6 illustrating such a dislocation in an infinite medium.
• Stress and displacement fields
For the dislocation depicted in Figure 2.6, the displacement component u3 is everywhere zero, as are all derivatives taken with respect to x3. Hence, we have a state of plane strain as described in Section 2.2. To this end, solving the elastic field reduces to solving the biharmonic Equation (2.20) with the appropriate boundary conditions. We summarize below the derivation of [26], which starts by setting
Φ =∇2Ψ. (2.31)
Then Φ satisfies Laplace equation. In polar coordinates, this means that ∂2
∂r2 + 1 r
∂
∂r + 1 r2
∂2
∂θ2
Φ = 0. (2.32)
It is well known that the harmonic Equation (2.32) is separable [26, 44] and thus
Figure 2.6: Edge dislocation along the positive x3 axis in a cartesian coordinate system.
has a single-valued solution of the form Φ =α0+β0lnr+
∞
X
n=1
αnrn+βnr−n
sinnθ+ γnrn+δnr−n
cosnθ . (2.33) Though we skip the details here, it can be shown using the appropriate boundary conditions that the simplest solution of the form (2.33) is
Φ =β1r−1sinθ , (2.34)
so that combining Equations (2.31) and (2.34) gives ∂2
∂r2 +1 r
∂
∂r + 1 r2
∂2
∂θ2
Ψ =β1r−1sinθ . (2.35) A solution of the above equation is
Ψ = β1
2rsinθlnr . (2.36)
With the requirement that a complete circuit around the dislocation must yield a displacement of b along the x1-direction, we find that
β1 =− µb π(1−ν).
Therefore, the solution of the biharmonic Equation (2.20) for the edge dislocation is
Ψ =− µb
2π(1−ν)rsinθlnr . (2.37) The nonzero components of the stress tensor follow as [19, 26, 35, 44]
σrr = 1 r2
∂2Ψ
∂θ2 +1 r
∂Ψ
∂r =− µb 2π(1−ν)
1
rsinθ , (2.38a) σθθ = ∂2Ψ
∂r2 =− µb 2π(1−ν)
1
rsinθ , (2.38b)
σrθ =σθr =− ∂
∂r 1
r
∂Ψ
∂θ
= µb
2π(1−ν) 1
rcosθ , (2.38c) σzz =ν(σrr+σθθ) = − µb
π(1−ν) ν
r sinθ . (2.38d)
The corresponding strains follow from Hooke’s law and the displacements from integration of the strains. The latter are found to be
ur = b 2π
− 1−2ν
2 (1−ν)sinθlnr+ sinθ
4 (1−ν) +θcosθ
, (2.39a)
uθθ = b 2π
− 1−2ν
2 (1−ν)cosθlnr− cosθ
4 (1−ν) −θsinθ
. (2.39b)
The equivalent counterparts in cartesian coordinates are Ψ =− µb
4π(1−ν)x2ln x21+x22
, (2.40)
σ11 = Ψ,22 =− µb 2π(1−ν)
x2(3x21+x22)
(x21+x22)2 , (2.41a) σ22 = Ψ,11 = µb
2π(1−ν)
x2(x21−x22)
(x21+x22)2 , (2.41b) σ12 =σ21 =−Ψ,12 = µb
2π(1−ν)
x1(x21−x22)
(x21+x22)2 , (2.41c) σ33 =ν(σ11+σ22) = − µbν
π(1−ν) x2
x21+x22, (2.41d) and
u1 = b 2π
tan−1
x2 x1
+ x1x2
2 (1−ν) (x21+x22)
, (2.42a)
u2 =− b 2π
1−2ν
4 (1−ν)ln x21+x22
+ x21−x22 4 (1−ν) (x21 +x22)
. (2.42b)
• Strain energy
For the same reasons as in the screw case, we will compute the elastic energy inside a hollow cylinder of inner radius r0 and outer radius R as depicted in Figure 2.7.
Using Equation (2.38) and Hooke’s law, we find W
L = ˆ R
r0
ˆ 2π 0
1
2{σrrεrr+σθθεθθ+σrθεrθ+σθrεθr}rdθdr
= ˆ R
r0
ˆ 2π 0
1 2
σrr1 +ν
E [(1−ν)σrr−νσθθ] +σθθ
1 +ν
E [(1−ν)σθθ−νσrr] + 2σrθ
σrθ 2µ
rdθdr
= µb2 4π(1−ν)ln
R r0
.
(2.43)
⇠
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x2
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✓
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(x<latexit sha1_base64="MwiIe5O5QNDF6Bk2G9fDJb+CuKM=">AAAB73icbVBNS8NAEJ3Ur1q/qh69LBahgpSkCOqt6MVjBWMrbQib7aZdupuE3Y20hP4KLx5UvPp3vPlv3LY5aOuDgcd7M8zMCxLOlLbtb6uwsrq2vlHcLG1t7+zulfcPHlScSkJdEvNYtgOsKGcRdTXTnLYTSbEIOG0Fw5up33qiUrE4utfjhHoC9yMWMoK1kR6rI985G/n1U79csWv2DGiZODmpQI6mX/7q9mKSChppwrFSHcdOtJdhqRnhdFLqpoommAxxn3YMjbCgystmB0/QiVF6KIylqUijmfp7IsNCqbEITKfAeqAWvan4n9dJdXjpZSxKUk0jMl8UphzpGE2/Rz0mKdF8bAgmkplbERlgiYk2GZVMCM7iy8vErdeuas7deaVxnadRhCM4hio4cAENuIUmuEBAwDO8wpslrRfr3fqYtxasfOYQ/sD6/AGuLI82</latexit><latexit sha1_base64="MwiIe5O5QNDF6Bk2G9fDJb+CuKM=">AAAB73icbVBNS8NAEJ3Ur1q/qh69LBahgpSkCOqt6MVjBWMrbQib7aZdupuE3Y20hP4KLx5UvPp3vPlv3LY5aOuDgcd7M8zMCxLOlLbtb6uwsrq2vlHcLG1t7+zulfcPHlScSkJdEvNYtgOsKGcRdTXTnLYTSbEIOG0Fw5up33qiUrE4utfjhHoC9yMWMoK1kR6rI985G/n1U79csWv2DGiZODmpQI6mX/7q9mKSChppwrFSHcdOtJdhqRnhdFLqpoommAxxn3YMjbCgystmB0/QiVF6KIylqUijmfp7IsNCqbEITKfAeqAWvan4n9dJdXjpZSxKUk0jMl8UphzpGE2/Rz0mKdF8bAgmkplbERlgiYk2GZVMCM7iy8vErdeuas7deaVxnadRhCM4hio4cAENuIUmuEBAwDO8wpslrRfr3fqYtxasfOYQ/sD6/AGuLI82</latexit><latexit sha1_base64="MwiIe5O5QNDF6Bk2G9fDJb+CuKM=">AAAB73icbVBNS8NAEJ3Ur1q/qh69LBahgpSkCOqt6MVjBWMrbQib7aZdupuE3Y20hP4KLx5UvPp3vPlv3LY5aOuDgcd7M8zMCxLOlLbtb6uwsrq2vlHcLG1t7+zulfcPHlScSkJdEvNYtgOsKGcRdTXTnLYTSbEIOG0Fw5up33qiUrE4utfjhHoC9yMWMoK1kR6rI985G/n1U79csWv2DGiZODmpQI6mX/7q9mKSChppwrFSHcdOtJdhqRnhdFLqpoommAxxn3YMjbCgystmB0/QiVF6KIylqUijmfp7IsNCqbEITKfAeqAWvan4n9dJdXjpZSxKUk0jMl8UphzpGE2/Rz0mKdF8bAgmkplbERlgiYk2GZVMCM7iy8vErdeuas7deaVxnadRhCM4hio4cAENuIUmuEBAwDO8wpslrRfr3fqYtxasfOYQ/sD6/AGuLI82</latexit><latexit sha1_base64="MwiIe5O5QNDF6Bk2G9fDJb+CuKM=">AAAB73icbVBNS8NAEJ3Ur1q/qh69LBahgpSkCOqt6MVjBWMrbQib7aZdupuE3Y20hP4KLx5UvPp3vPlv3LY5aOuDgcd7M8zMCxLOlLbtb6uwsrq2vlHcLG1t7+zulfcPHlScSkJdEvNYtgOsKGcRdTXTnLYTSbEIOG0Fw5up33qiUrE4utfjhHoC9yMWMoK1kR6rI985G/n1U79csWv2DGiZODmpQI6mX/7q9mKSChppwrFSHcdOtJdhqRnhdFLqpoommAxxn3YMjbCgystmB0/QiVF6KIylqUijmfp7IsNCqbEITKfAeqAWvan4n9dJdXjpZSxKUk0jMl8UphzpGE2/Rz0mKdF8bAgmkplbERlgiYk2GZVMCM7iy8vErdeuas7deaVxnadRhCM4hio4cAENuIUmuEBAwDO8wpslrRfr3fqYtxasfOYQ/sD6/AGuLI82</latexit> 1, x2)
r
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r0
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R
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Figure 2.7: Edge dislocation along the positive x3 axis in a hollow cylinder used to compute the strain energy of the dislocation.
General straight dislocations
We now turn to the treatment of a general straight dislocation. In doing so, we remember that the Burgers vector of such a dislocation can always be decomposed into edge and screw components—see Figure 2.8.
• Stress and displacement fields
Recall from Equation (1.6) that the screw component of the dislocation is bs = (b·ξ)ξ,